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3g urea is dissolved in 45g of water. The relative lowering of vapour pressure is(mol .wt. of urea=60)

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0.02

0.01

0.05

0.04

answer is D.

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Detailed Solution

R.L.V.P = $\frac{n_{2}}{n_{1} + n_{2}}$

$n_{2} = \frac{3}{60} = 0.05$ , $n_{1} = \frac{45}{18} = 2.5$

R.L.V.P = $\frac{0.05}{2.5 + 0.05} = 0.0196 = 0.02$

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